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Chapter 10: Problem 71

Draw the Lewis structure for the \(\mathrm{BeCl}_{4}^{2-}\) ion. Predict its geometry and describe the hybridization state of the Be atom.

Short Answer

Expert verified
The Lewis structure of \(\mathrm{BeCl}_{4}^{2-}\) has a Be atom in the center surrounded by 4 Cl atoms. Each Be-Cl has a single bond and the Be atom carries two extra electrons. The molecule has a tetrahedral geometry. The Be atom's hybridization state is sp3d.

Step by step solution

01

Count Valence Electrons

The atom Be has 2 valence electrons and each atom Cl has 7 valence electrons. In \(\mathrm{BeCl}_{4}^{2-}\), there are 4 Cl atoms. There are also 2 additional electrons because of the '2-' charge. Thus, we have \(2 + (4 \cdot 7) + 2 = 30 \) valence electrons.
02

Draw the Lewis Structure

Berillium (Be) as the least electronegative atom becomes the central atom. Leaving Be in the center, organize the 4 Cl atoms around it. Be, being in the 2nd period of the periodic table, can expand its octet to bond with the 4 Cl atoms. Draw a single bond between the Be atom and each Cl atom using a pair of electrons. Each Cl atom still has 6 valence electrons left, place them around each Cl atom (3 lone pairs around each). The total electrons used now equals \(4\cdot 2 + 3\cdot 6\cdot 4 = 28\), which is less than 30. Hence, place the two remaining electrons on Be.
03

Predict the Geometry

Apply the VSEPR theory to predict the geometry. Be has 4 bond pairs and no lone pair, hence the geometry will be tetrahedral according to VSEPR.
04

Describe the Hybridization State

Hybridization state can be calculated using the hybridization trigonal formula which is number of bond pairs + number of lone pairs on the central atom. In this case therefore, there are 4 bond pairs and 1 lone pair on the central Be atom. This adds up to 5, which implies sp3d hybridization (s represents one orbital, and each p and d represent three and five orbitals respectively, thus total five orbitals imply sp3d hybridization).

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